Question

Please help A student increases the temperature of a 579cm^3 balloon from 215 K to 575 K Assuming constant pressure, what should the new volume of the balloon be? Round your answer to one decimal place.

Answer 1

1548.5cubic centimeters.

Explanations:

According to Charles law, the volume of a given mass is directly proportional to the temperature provided that the pressure is constant. Mathematically;

`\begin{gathered} V\alpha T \\ V=kT \\ k=(V_1)/(T_1)=(V_2)/(T_2) \end{gathered}`

where

V1 and V2 are the volumes

T1 and T2 are the temperatures

Given the following parameters

`\begin{gathered} V_1=579cm^3 \\ T_1=215K \\ T_2=575K \end{gathered}`

Required

New volume V2

Substitute

`\begin{gathered} V_2=(V_1T_2)/(T_1) \\ V_2=(579*575)/(215) \\ V_2=(332925)/(215) \\ V_2=1548.48cm^3 \end{gathered}`

Hence the new volume rounded to nearest one decimal place is 1548.5cubic centimeters.